Question

Given f(x,y)=x^2+4y^2. using x=1 and y=1 and differentials, find an approximation for (1.01)^2 + 4(1.99)^2

Answer 1

TRY USING WOLFRAME ALPHA

Answer 2

ok i have it x0=1 deltax=0.01 f(x0+deltax)~f(x0)+dy 1.01^2~1+2*1*0.01=1.02
the second part y0=2 deltay=0.01 so 4y^2=4*1.99^2=16-8*2*0.01=15.84 , so the answer is 16.86

Answer 3

two morons